Question

The equilibrium constant for the following reaction has been estimated to be K = 1020 at 25∘C.

NH−2(aq)+H2O(l)⇋NH3(aq)+OH−(l)

You may want to reference (Page) Section 15.1 while completing this problem.

Part A

Estimate the equilibrium concentration of NH−2 in a solution prepared by dissolving 0.0129 mol NaNH2 in water to make 1.00 L of solution at 25∘C.

Express your answer using three significant figures.

Answer 1

Molarity of NH2- = MOL / volume of solution in litres

= 0.0129 mol / 1 L

= 0.0129 M

NH2- ...+..... H2O <------>........ NH3 + OH-

0.0129...... ..................................0........0.. (INITIAL)

-x . . ....... . . .................................. +x......+x.. (CHANGE)

0.0129 - x .. ..... ............... ..........+x...... +x..... (AT EQUILIBRIUM)

K = [NH3] [OH-] / [NH2-]

1020 = x^2 / ( 0.0129 - x)

x^2 + 1020 x - 13.158 = 0

solve the equation for x

x = 0.0128998

So, at equilibrium [NH2-] = 0.0129 - x

= 0.0129 - 0.0128

= 1.00 x 10^-4........... (3 significant figures)